01 May 05
The vertices A and C of a triangle ABC lie on the circumference of a circle. The circle cuts the sides AB and CB at points X and Y respectively. IF AC=XC and Angle YXC = 40 degrees, find the size of Angle ABC.
& (note these are 2 different questions)
In a triangle ABC, Angle ABC = 112 degrees and AB = BC. D is a point on the side AB and E is a point on the side AC such that AE = ED = DB. Find the size of Angle BCD.
02 May 05
Originally posted by phgaoBy my calculation, I get that ang(ABC) should be 45 degrees.
The vertices A and C of a triangle ABC lie on the circumference of a circle. The circle cuts the sides AB and CB at points X and Y respectively. IF AC=XC and Angle YXC = 40 degrees, find the size of Angle ABC.
02 May 05
2 edits
Originally posted by phgaohmmm...actually, i think i made a mistake earlier. I think ABC should be 40 for the first problem.
Can you provide proof? And what about the 2nd Q?
My proof of the first question (I think this is right):
We are given that AC = XC. Therefore, it follows (law of sines, for example) that CAX = AXC. (Here my notation is that CAX refers to angle CAX)
So let CAX = AXC = x.
Also, let the radius of the circle be R.
It follows from the law of sines that XC/[sin(x)] = 2R = XC/[sin(XYC)].
It also follows from law of sines that AY/[sin(x+40)] = 2R = AY/[sin(ACY)].
We conclude from these two that:
sin(x) = sin(XYC) and
sin(x+40) = sin(ACY).
This does not mean that x = XYC and x+40 = ACY. In fact, these cannot both hold because then ABC would be 0 which doesn't make sense.
So the only solution I see that is in keeping with YXC = 40 is
XYC = 180-x and ACY = x+40. (for x < 90).
Then to find x, we just set the sum of angles within the circle to 360:
XAC + ACY + XYC + YXA = 360
-> x + (x+40) + (180-x) + (x+40) = 360
-> x = 50
Then it follows that ABC = 180 - XAC - ACY
-> ABC = 180 - (X) - (X+40) = 180 - 50 - 90 = 40
For the second problem, I was getting confused...I kept getting that the problem was not constrained, but I take it that it is, so I need to look at that problem again...
02 May 05
4 edits
Originally posted by phgaoOkay, for the second problem, here is what I get, although I don't have a calculator handy, so I haven't simplified it to the numerical answer:
In a triangle ABC, Angle ABC = 112 degrees and AB = BC. D is a point on the side AB and E is a point on the side AC such that AE = ED = DB. Find the size of Angle BCD.
We are given that ABC = 112, and AB = BC. We are also given that AE = ED = DB.
From these, we know that BDC = (68 - BCD)
So we invoke the Law of Sines:
BC/[sin(68 - BCD)] = DB/[sin(BCD)]
Also, AB/[sin(AEB = 129)] = AE/[sin(ABE = 17)].
Then noting that DB = AE and AB = BC, we can simplify these two equations to solve for BCD. When I do that, I get:
cot(BCD) = [(sin(129))/(sin(68)sin(17))] + cot(68)
I don't have a calculator to see what number I get.
Does this look right to you?
02 May 05
Originally posted by davegageNow that I have my trusty calculator,
Okay, for the second problem, here is what I get, although I don't have a calculator handy, so I haven't simplified it to the numerical answer:
We are given that ABC = 112, and AB = BC. We are also given that AE = ED = DB.
From these, we know that BDC = (68 - BCD)
So we invoke the Law of Sines:
BC/[sin(68 - BCD)] = DB/[sin(BCD)]
Also, AB/ ...[text shortened]... cot(68)
I don't have a calculator to see what number I get.
Does this look right to you?
I get that for Question 2, BCD = 17.